- Use contour integration to show that
\[ \int_ {- \infty} ^ {\infty} \frac {e ^ {- 2 \pi i x \xi}}{(1 + x ^ {2}) ^ {2}} d x = \frac {\pi}{2} (1 + 2 \pi | \xi |) e ^ {- 2 \pi | \xi |} \]
for all ξ real.
Solution
Let \[ I(\xi)=\int_{-\infty}^{\infty}\frac{e^{-2\pi i x\xi}}{(1+x^2)^2}\,dx. \] First assume \(\xi>0\). Close the contour in the lower half-plane, where \[ |e^{-2\pi i z\xi}|=e^{2\pi\xi\operatorname{Im}z} \] decays. The only enclosed pole is the double pole at \(z=-i\). Since the contour is clockwise, \[ I(\xi)=-2\pi i\operatorname{Res}_{z=-i} \frac{e^{-2\pi i z\xi}}{(z-i)^2(z+i)^2}. \] The residue is \[ \left.\frac{d}{dz}\frac{e^{-2\pi i z\xi}}{(z-i)^2}\right|_{z=-i} =e^{-2\pi\xi} \left(\frac{-2\pi i\xi}{(-2i)^2}-\frac{2}{(-2i)^3}\right) =\frac{i}{4}(1+2\pi\xi)e^{-2\pi\xi}. \] Therefore \[ I(\xi)=\frac\pi2(1+2\pi\xi)e^{-2\pi\xi} \qquad(\xi>0). \]
For \(\xi<0\), close in the upper half-plane; the analogous residue at \(z=i\) gives \[ I(\xi)=\frac\pi2(1-2\pi\xi)e^{2\pi\xi} =\frac\pi2(1+2\pi|\xi|)e^{-2\pi|\xi|}. \] At \(\xi=0\), either evaluate directly or let \(\xi\to0\); in both cases \(I(0)=\pi/2\). Thus for every real \(\xi\), \[ \boxed{I(\xi)=\frac\pi2(1+2\pi|\xi|)e^{-2\pi|\xi|}}. \]